Cylinder and cones (Putnam Exam 1938) Right circular cones of height h and radius r are attached to each end of a right circular cylinder of height h and radius r, forming a double-pointed object. For a given surface area A, what are the dimensions r and h that maximize the volume of the object?

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1. Define the total surface area A of the double-pointed object, which includes the surface area of the cylinder and the two cones. The surface area of the cylinder is given by 2πrh (lateral area) plus the area of the two circular bases, which is 2πr². The surface area of one cone is πr√(r² + h²) (lateral area) plus the base area πr², so for two cones, it is 2(πr√(r² + h²) + πr²). Combine these to express A in terms of r and h.
2. Set up the equation for the total surface area A: A = 2πrh + 2πr² + 2πr√(r² + h²).
3. Express the volume V of the object as a function of r and h. The volume of the cylinder is V_cylinder = πr²h, and the volume of one cone is V_cone = (1/3)πr²h. Therefore, the total volume V = πr²h + 2(1/3)πr²h = πr²h + (2/3)πr²h = (5/3)πr²h.
4. Use the method of Lagrange multipliers or substitute h in terms of r from the surface area equation into the volume equation to express V solely in terms of r.
5. Differentiate the volume function with respect to r, set the derivative equal to zero to find critical points, and analyze these points to determine the dimensions r and h that maximize the volume.